Answer to Question #190065 in Inorganic Chemistry for julieanna robinson

Question #190065
  1. Consider the following reaction: 2 Al + 6 HBr → 2 AlBr3 + 3 H2 


  1. What is the limiting reactant when 3.22 moles of Al reacts with 4.96 moles of HBr? 
  2. How many moles of H2 are formed?
1
Expert's answer
2021-05-07T05:25:39-0400

1) 2Al +1)\ 2Al \ +6HBr6HBr\to 2AlBr3+3H22AlBr_3+3H_2

Since 2 moles aluminium are required for every 6 moles of hydrogen bromide, yet we only have 3.22 moles aluminium and 4.96 moles of hydrogen bromide , it implies that HBrr will get used up first and hence it is limiting reactant while Al is in excess.


So all 4.96 mol of HBr is used up and reacts with =4.963=1.653 molAl\ =\dfrac{4.96}{3}=1.653 \ mol Al to produce=4.962=2.48 molH2=\dfrac{4.96}{2}=2.48\ mol H_2


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